The ratio of nucleus density of nitrogen nucleus $({}_7^{14}N)$ to the nucleus density of oxygen nucleus $({}_8^{16}O)$ will be:

1 : 1

The correct answer is Option (2) → 1 : 1

R, Radius of nucleus $∝ A^{1/3}$ [A = mass number]

V, Volume of nucleus $∝ A^{1/3})^3$

$∝ A$

$∴ P, Density = \frac{Mass}{Volume}=\frac{A}{A}=A^0$

$∴\frac{P_N}{P_O}=\frac{R_OA^0}{R_OA^0}=\frac{1}{1}$